Question

solve the equation for t, 0 less than or equal to t < 2pi
2cot(t)-csc(t)=3
choices are
{1.18}
{0.87,3.45}
{0.87,3.14,3.45}
{0.84,4.24}

Answer 1

Simple cot(t)= cos(t)/sin(t)
Csc(t)=1/sint
2cos(t)-1=3sin(t)
Cos(t)=sqrt(1-(sin(t))^2))
1-(sint)^2=((3sint+1)/2))^2
4-4(sint)^2=9(sint)^2+6sint+1
5(sint)^2+6sint-3=0
now you have a quadratic in terms of sine t solve it you would get two values for sin t now since the value of sint 0<=Sin t<=1 there for any value less than zero and any value greater than zero would be discarded

Answer 2

now use the formula for quadratics to obtain the root becuase this equation cannot be factorised

Answer 3

Above should be 13 sin ^2 t not 5 sin^2 t.

Anyway, I think there is something wrong with the question...

Answer 4

@estudier the question is correct look at my simplification

Answer 5

Your simplification contains the above error.

But either way, the answer choices or the question is wrong.

Answer 6

@estudier right it would be 13

Answer 7

after I put it in quadratic equation I get
\[\frac{ -6\pm \sqrt{6^{2}-4(13)(-3)} }{ 2(13) }\]
then I simplify to
\[\frac{ -6\pm 8\sqrt{3} }{ 26 }\]

Answer 8

that leaves me with
\[\frac{ -3\pm 4\sqrt{3} }{ 13 }\]

Answer 9

put 4sqrt(3)=4x1.732